= Solution
The <divergence> is $y^2+z^2+x^2$. Over the cube $[-1/2,1/2]^3$, each squared-coordinate term has integral $\int_{-1/2}^{1/2}s^2ds=1/12$, while the other two coordinate integrals have length one. Thus
$$
\boxed{\int_V\nabla\cdot F\,dV=3\cdot\frac1{12}=\frac14.}
$$
Compute the surface <flux> separately. On $x=1/2$, the outward normal is $e_x$ and $F\cdot n=y^2/2$, giving flux $1/24$. On $x=-1/2$, the normal is $-e_x$, so $F\cdot n=-xy^2=y^2/2$, again giving $1/24$. The two $y$ faces similarly give $z^2/2$ each, and the two $z$ faces give $x^2/2$ each. All six face integrals therefore give
$$
\boxed{\int_{\partial V}F\cdot dS=6\cdot\frac1{24}=\frac14.}
$$
The independently evaluated values agree, verifying the <divergence theorem> for this field and cube.
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